Python | Filter String with substring at specific position

Sometimes, while working with Python string lists, we can have a problem in which we need to extract only those lists that have a specific substring at a specific position. This kind of problem can come in data processing and web development domains. Let us discuss certain ways in which this task can be performed. 

Method #1: Using list comprehension + list slicing The combination of the above functionalities can be used to perform this particular task. In this, we check for substring range using list slicing, and the task of extraction list is compiled in list comprehension. 

# Python3 code to demonstrate
# Filter String with substring at specific position
# using list comprehension + list slicing

# Initializing list
test_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks']

# printing original list
print("The original list is : " + str(test_list))

# Initializing substring
sub_str = 'geeks'

# Initializing range
i, j = 0, 5

# Filter String with substring at specific position
# using list comprehension + list slicing
res = [ele for ele in test_list if ele[i: j] == sub_str]

# printing result
print("Filtered list : " + str(res))

The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks']
Filtered list : ['geeksforgeeks', 'geeks']

Time Complexity: O(n) where n is the total number of values in the list “test_list”. 
Auxiliary Space: O(n) where n is the total number of values in the list “test_list”.

Method #2: Using filter() + lambda The combination of the above methods can be used to perform this task. In this, we filter the elements using logic compiled using lambda using filter(). 

# Python3 code to demonstrate
# Filter String with substring at specific position
# using filter() + lambda

# Initializing list
test_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks']

# printing original list
print("The original list is : " + str(test_list))

# Initializing substring
sub_str = 'geeks'

# Initializing range
i, j = 0, 5

# Filter String with substring at specific position
# using filter() + lambda
res = list(filter(lambda ele: ele[i: j] == sub_str, test_list))

# printing result
print("Filtered list : " + str(res))

The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks']
Filtered list : ['geeksforgeeks', 'geeks']

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Auxiliary Space: O(n)

Method #3 : Using find() method

# Python3 code to demonstrate
# Filter String with substring at specific position

# Initializing list
test_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks']

# printing original list
print("The original list is : " + str(test_list))

# Initializing substring
sub_str = 'geeks'

# Initializing range
i, j = 0, 5

# Filter String with substring at specific position
res=[]
for k in test_list:
    if k.find(sub_str)==i:
        res.append(k)
# printing result
print ("Filtered list : " + str(res))

The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks']
Filtered list : ['geeksforgeeks', 'geeks']

Time Complexity: O(n), where n is the length of the input list. This is because we’re using the built-in  find() method which has a time complexity of O(n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list. 

Method 4 :  Using a for loop and string slicing

step by step approach for the program:

1. Initialize the list test_list with some strings.
2. Print the original list using the print() function.
3. Initialize a substring sub_str which we want to search for in the list of strings.
4. Initialize two variables i and j which represent the range in which we want to check for the substring. In this case, we want to check if the substring exists at the start of the string, so i is 0 and j is 5 (length of sub_str).
5. Create an empty list res to store the filtered strings.
6. Loop through each string k in test_list.
7. Check if the substring sub_str exists at the specific position using string slicing. If it does, append the string k to the res list.
8. Print the filtered list using the print() function.

## Python3 code to demonstrate
# Filter String with substring at specific position

# Initializing list
test_list = ['geeksforgeeks', 'is', 'best', 'for', 'geeks']

# printing original list
print("The original list is : " + str(test_list))

# Initializing substring
sub_str = 'geeks'

# Initializing range
i, j = 0, 5

# Filter String with substring at specific position
res=[]
for k in test_list:
    if k[i:i+len(sub_str)] == sub_str:
        res.append(k)

# printing result
print ("Filtered list : " + str(res))

The original list is : ['geeksforgeeks', 'is', 'best', 'for', 'geeks']
Filtered list : ['geeksforgeeks', 'geeks']

The time complexity of this program is O(n*m), where n is the length of the test_list and m is the length of the sub_str.

The auxiliary space of this program is O(k), where k is the size of the resulting list that contains the filtered strings. 

manjeet_04

Follow

Improve

Article Tags :

• Python
• Python Programs
• Python list-programs
• Python string-programs

Practice Tags :

• python